CODEWITHSUNDEEP
pythonfunction
Viktor Berg
Viktor Berg

Reputation: 15

calling functions python from outside

I have small a python script looking something like this:

def number1():
    x = 1
    open_numbers = []
    open_numbers.append(x)
    return open_numbers

def myfunction(open_numbers):
    y = 1
    open_numbers.append(y)

I would like to call the myfunction in the the end of the script. Using

myfunction()

But it keeps telling me missing 1 required positional argument: 'open_numbers'

Tried passing the argument and got name 'open_numbers' is not defined

I plan to add more functions later and run them the same way

function(arg)
function2(arg)
function3(arg)

Upvotes: 0

Views: 158

Answers (3)

CypherX
CypherX

Reputation: 7353

Solution

First of all, your code was not properly indented. I have corrected that.
The function myfunction takes in a list (open_numbers) as input and should return it as well.

I have passed in the output of number1() as the input to myfunction(). This should create a list: [1, 1]. And that's what it did. 

def number1():
    x = 1
    open_numbers = []
    open_numbers.append(x)
    return open_numbers


def myfunction(open_numbers):
    y = 1
    open_numbers.append(y)
    return open_numbers

myfunction(number1())

Output: 

[1, 1]

Upvotes: 1

Jacek Rojek
Jacek Rojek

Reputation: 1122

You might want to define default parameter, and return the updated value

def myfunction(open_numbers = []):
    y = 1
    open_numbers.append(y)
    return open_numbers

Then you can call it with passing parameter myfunction([1]) or without myfunction()

Upvotes: 0

Sadrach Pierre
Sadrach Pierre

Reputation: 121

you need to pass in an object to your function. you can call your function with an empty list if you want:

a = []
myfunction(a)

Upvotes: 1

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